Part 3 · Light in matter

3.4 Dispersion and group velocity

Source: Postorino, week 4 slides, p. 11, and week 3 slides, p. 11

Objectives

Theorem. In a medium with index $n (ω)$ , $v_{g} = \frac{c}{n + ω d n / d ω}$ .
Picture. The carrier moves at $v_{f} = c / n$ , the envelope (the signal, the energy) at $v_{g}$ ; with normal dispersion pulses are slower than crests.
Key. As $ω \to \infty$ , $n \to 1$ : one polarisation mechanism after another is frozen out.

Carrier and envelope, again

Back to beats (Proposition 2.4.5): $E (x, t) = 2 E_{0} envelope (modulation) cos (Δ k \cdot x - Δ ω \cdot t) \cdot carrier wave cos (k_{0} x - ω_{0} t) .$ The signal has two parts:

the carrier $cos (k_{0} x - ω_{0} t)$ oscillates rapidly and moves at the phase velocity $v_{f} = ω_{0} / k_{0}$ ;
the envelope $cos (Δ k \cdot x - Δ ω \cdot t)$ modulates the amplitude, the “shape” of the signal, and moves at $v_{g} = \frac{Δ ω}{Δ k} \to v_{g} = \frac{d ω}{d k} in the limit of a continuum of frequencies (Δ k \to 0) .$

(Week 4 writes the phase velocity $v_{f}$ ; week 3 wrote $v_{φ}$ . Here $ω_{0}$ and $k_{0}$ are the carrier’s, not a resonance.)

Group velocity in a medium

Theorem 3.4.1Group velocity in a dispersive mediumslides w4, p. 11

In a medium with $n = n (ω)$ (Theorem 3.2.5), $v_{g} = \frac{c}{n ( ω ) + ω \frac{d n}{d ω}} .$

Proof

$k (ω) = \frac{ω}{v _{f}} = \frac{ω n ( ω )}{c}$ , so $\frac{d k}{d ω} = \frac{d}{d ω} (\frac{ω n ( ω )}{c}) = \frac{1}{c} (n (ω) + ω \frac{d n}{d ω}),$ and $v_{g} = d ω / d k = 1/ (d k / d ω)$ .

Key observation 3.4.2How far v_g is from v_fslides w4, p. 11

How far $v_{g}$ is from $v_{f} = c / n$ depends on the product of $d n / d ω$ with $ω$ . With no dispersion, $d n / d ω = 0$ and the two are equal.

beyond the lectures

With normal dispersion, $d n / d ω > 0$ (glass in the visible), $v_{g} < v_{f}$ : pulses travel slower than the crests.
Inside an absorption line, $d n / d ω < 0$ and the formula can even give $v_{g} > c$ . There the pulse is so strongly absorbed and reshaped that “group velocity” no longer describes the speed of energy or signals. Nothing travels faster than light.

Loading the demo…

Figure 3.4.1 A modulated wave in glass, where normal dispersion makes the group slower than the crests. Press Play: the

crest

marker overtakes the

envelope

marker. Push

v_{g} / v_{φ}

above 1 for the anomalous case.

Example 3.4.3A pulse in glassbeyond the lectures

In glass near 500 nm, $n = 1.52$ and $d n / d λ = - 5 \times 1 0^{- 5}$ nm⁻¹. Since $ω = 2 π c / λ$ , $ω \frac{d n}{d ω} = - λ \frac{d n}{d λ} = 0.025$ . So $v_{g} = c /1.545 = 0.647 c$ , while $v_{f} = c /1.52 = 0.658 c$ : the pulse is about 2 % slower than its crests.

Frozen out

Key observation 3.4.4At high frequency matter is empty spaceslides w4, p. 11

N.B. As $ω \to \infty$ , $n (ω) \to 1$ : for $λ \to 0$ matter is mostly empty space (Sizes and masses). At very high frequency every polarisation mechanism is too slow to follow. The degrees of freedom are frozen out, one after another as the frequency rises, as in the steps of $n$ across $ω_{1}, ω_{2}, ω_{3}$ (Figure 3.2.3). Compare the specific heat of diatomic molecules, where rotations and vibrations switch off at low temperature.

The same word comes back for the blackbody: a mode whose energy quantum is much larger than $K_{B} T$ is frozen out (§5.2 Rayleigh–Jeans and Planck).

Exercises

Exercise 3.4.1beyond the lectures

Derive the group-velocity formula a second way: differentiate $ω n (ω) = c k$ with respect to $k$ .

Solution

$\frac{d}{d k} [ωn (ω)] = (n + ω \frac{d n}{d ω}) \frac{d ω}{d k} = c$ , so $v_{g} = \frac{d ω}{d k} = \frac{c}{n + ω d n / d ω}$ .

Exercise 3.4.2beyond the lectures

For a Drude metal above the plasma frequency (ignore collisions), $ω^{2} = ω_{P}^{2} + c^{2} k^{2}$ . Find $v_{f}$ and $v_{g}$ and show $v_{f} v_{g} = c^{2}$ .

Solution

$v_{f} = ω / k = ω_{P}^{2} + c^{2} k^{2} / k > c$ ; $v_{g} = d ω / d k = c^{2} k / ω < c$ . Their product is $c^{2}$ . The phase velocity exceeds $c$ but carries no signal; the group velocity, which does, stays below $c$ .

Exercise 3.4.3beyond the lectures

Why does a prism need normal dispersion to send blue further than red?

Solution

Snell’s law bends a ray more for larger $n$ . With normal dispersion $n$ grows with frequency, so blue (higher $ω$ ) has the larger $n$ and is deviated more.